20. Electrochemistry

Question 1

What is electrolysis?

Answer 1

The decomposition of a compound into its elements by an electric current. Carried out in an electrolysis cell.

Question 2

What can be produced from electrolysis?

Answer 2

Reactive metals can be extracted from their ores (too high in reactivity series to be extracted by heating with carbon). Metals formed are either deposited as a layer on the cathode or form a molten layer in the cell.
Non-metals like chlorine can be produced. Gases are bubbled off.
Metals can be purified.

Question 3

What is an electrolyte?

Answer 3

The compound that is to be decomposed. It can be a molten ionic compound or a conc. aqueous solution of ions.

Question 4

What is an electrode?

Answer 4

Rods made from either graphite or a metal, used to conduct electricity to and from the electrolyte.
Anode = positive, cathode = negative.

Question 5

What is the power supply in an electrolysis cell?

Answer 5

Direct current.

Question 6

The mass deposited at an electrode is proportional to…

Answer 6

Time over which an electric current passes, and the strength of the current.

AKA the quantity of electricity passing through the electrolyte, given by Q = It.

Question 7

What is a Faraday?

Answer 7

The quantity of electric charge carried by 1 mole of electrons or singly charged ions - 96500C/mol.

Question 8

How is the Avogadro constant related to the Faraday?

Answer 8

F = Le

• F = charge on 1 mole of electrons
• e = charge on 1 electron (1.6 x 10^-19)

Question 9

What is the electrolytic method for finding the charge on an electron?

Answer 9

Electrolysis cell with Cu cathode and anode. Ammeter, d.c. power supply and variable resistor (maintains a constant current) connected in series.

• Weigh the electrodes separately.
• Apply a constant electric current for a measured time period.
• Wash and dry the electrodes (distilled water, then propanone) and re-weigh.
• The cathode should have increased in mass, the anode should have decreased - measure the loss in mass.
• Calculation should support F = Le.

Question 10

When does a redox EQM exist?

Answer 10

Between chemically related species in different oxidation states, eg. copper rod in solution of its ions. It is established when the rate of electron gain = rate of electron loss.

There are two opposing reactions:

• metal atoms from the rod entering the solution as metal ions, leaving electrons behind on the surface of the rod
• copper ions in solution accepting electrons from the rod, deposited on the surface of the rod as copper metal.

Question 11

What does reactivity have to do with redox EQM?

Answer 11

For unreactive metals, EQM lies further to the right (compared with other metal EQM). For reactive metals, it lies further to the left (eg. V 2+ ions).

The more reactive, the harder it is to reduce, so the further the EQM will lie to the left.

Question 12

What is electrode/reduction potential?

Answer 12

When a metal is placed into a solution of its ions, a voltage (electric potential) is set up between the two. This cannot be measured directly but the difference between this and another system can be (uses the standard hydrogen electrode).

Question 13

How is this electric potential caused?

Answer 13

Through the formation of an electrical double layer when an element is placed in a solution of its ions, eg. Zn and Zn2+.

• some atoms on Zn metal’s surface release electrons and are converted to Zn 2+, which go into solution.
• excess electrons are left on Zn’s surface - these attract the excess Zn 2+ ions that are in the surrounding solution.
• a double layer is formed - this charge buildup causes a voltage between the metal atoms and ions in solution.

Question 14

Describe the standard hydrogen electrode.

Answer 14

• H+ ions in solution (1.00 mol/dm3)
• H2 gas at 1 ATM pressure (forms EQM with H+)
• inert platinum electrode covered with platinum black (finely divided Pt) in contact with the EQM.

The finely divided Pt allows for H2 and H+ to be in contact and set up an EQM quickly.
Standard electrode potentials for all half-cells are measured against this - its own EΘ is 0.00V.

Question 15

What does an electrode potential tell us?

Answer 15

How easy it is to reduce a substance. It refers to the reduction reaction, so electrons always appear on the LEFT. More positive = easier to reduce species on the left. Metal on right is relatively unreactive / poor reducing agent (for the other half-cell).

Question 16

What is standard electrode potential?

Answer 16

For a half-cell; the voltage measured under standard conditions with the standard hydrogen electrode as the other half-cell.
It is affected by temperature, pressure and concentration so STD conditions must be used.

Question 17

Describe a metal/metal ion half-cell.

Answer 17

Metal in contact with a solution of its ions, eg. Cu2+/Cu, EΘ +0.34V.
- It is the positive pole - Cu2+ is easier to reduce than H+ (more likely to accept electrons, so it will gain electrons from H+/H2 half-cell and the H+/H2 half-cell will lose electrons to the Cu2+/Cu half-cell).

Question 18

Describe a nonmetal/nonmetal ion half-cell.

Answer 18

Nonmetal in contact with a solution of its ions, eg. Cl2/Cl-, EΘ +1.36V.

• It is the positive pole; H+/H2 is the negative pole.
• uses Pt wire/foil as electrode, surface of which EQM is established. It is in contact with the element and the aqueous solution of its ions.

Question 19

Describe an ion/ion half-cell.

Answer 19

They have ions of the same element with different oxidation states, eg. Fe3+/Fe2+, EΘ +0/77V.
- Pt or graphite electrode.

eg2. MnO4- + 8H+ + 5e- -> Mn2+ + 4H2O.
- H+ needed for the reduction reaction so all three ions are present in solution (1mol/dm3 each).

Question 20

What is the standard cell potential?

Answer 20

The difference in EΘ values between two half-cells in an electrochemical cell.
The less positive is always subtracted from the more positive.

Question 21

How can you predict the direction of electron flow?

Answer 21

eg. Zn2+/Zn (EΘ -0.76V), Ag+/Ag (EΘ +0.80V).
- Zn metal will lose electrons to the Ag+/Ag half-cell - these will travel through wires in the external circuit to this half-cell to reduce Ag+.

Electrons will flow from the negative pole to the positive pole.

Question 22

How can you predict the occurrence of a reaction?

Answer 22

In half-equations, the more oxidised species is on the left (more easily reduced), and vice versa.
EΘ can predict whether the reaction will proceed in its forward (reduction) direction or its reverse (oxidation) reaction.

Question 23

What is feasibility?

Answer 23

The feasibility is how likely a reaction is to occur. According to the ionic equation:
Zn + Cu2+ -> Zn2+ + Cu, we can see that placing Zn metal in 1M Cu2+ solution is feasible, but doing the reverse is not.
Using the reduction potential list, we can see the direction of reaction can be given by a clockwise pattern (reactant, product, reactant, product).

Question 24

How do you calculate the occurrence of a reaction?

Answer 24

1) Write the half-equations.
2) Identify the stronger oxidising and reducing agents.
3) The stronger O agent will react with the stronger R agent in a feasible reaction.
4) Combine the two half-equations.

Question 25

How do you find the relative voltage without knowing the better O/R agents?

Answer 25

1) Write the proposed reaction. 2) Write the corresponding half-equations. 3) Find EΘ for each reaction, flip the sign for the reaction that is reversed to oxidation. 4) Add the two voltages (we are combining the half-equations). 5) If the voltage is positive then the reaction is feasible.

Question 26

As EΘ values for reduction reactions get less positive...

Answer 26

The species on the left become weaker oxidising agents (accept electrons less readily) and the species on the right become stronger reducing agents (release electrons more readily).

Question 27

How do you reduce/oxidise a substance with a very low/high EΘ value?

Answer 27

eg. Zn2+ must be reacted with a stronger R agent, eg. Mg2+/Mg half-cell. eg2. Cl- won't be oxidised by HNO3 - it must be reacted with acidified MnO4-.

Question 28

Describe the trend in Group 17 as oxidising agents.

Answer 28

As you go down the group, the reduction potentials get less positive. The halogens are less easily reduced and their oxidising power decreases. The halides' reducing power will therefore increase as they are more easily oxidised going down the group.

Question 29

What happens to EΘ under non-standard conditions?

Answer 29

When we compare the voltage of STD half-cell X with STD hydrogen electrode, we are measuring the EΘ value of X. If T, P or c change, the conditions are non-STD and EΘ changes to E.

Question 30

How does ion concentration affect redox equilibria?

Answer 30

eg. Zn2+ + 2e- -> Zn - if [Zn2+] increases > 1M, E becomes less negative. - if [Zn2+] decreases < 1M, E becomes more negative. eg. Fe3+ + e- -> Fe2+ - if [Fe3+] > 1M, E increases - if [Fe3+] < 1M, E decreases - if [Fe2+] > 1M, E decreases. - if [Fe2+] < 1M, E increases. Increasing [Fe3+] and [Fe2+] will cancel each other out. Under non-STD conditions, if the EΘ values of the half-reactions involved have a difference >0.30V, the predicted reaction is highly feasible.

Question 31

Do feasibility predictions always work?

Answer 31

They only tell us if a reaction is possible and that it won't occur in the reverse direction, but they don't tell us anything about the rate. Sometimes the rate of reaction can be so slow that it seems as if nothing is happening. eg. Zn in cold water: - H+ + e- -> 1/2 H2 (EΘ 0.00V) - Zn2+ + 2e- -> Zn (EΘ -0.76V) Even considering low [H+], the EΘ values show that the reaction should occur, but it is very slow. Therefore, the rate of reaction determines LACK OF REACTIVITY rather than the EΘ.

Question 32

What does the Nernst equation tell us?

Answer 32

The effect of concentration and temperature on Ecell.

Question 33

State the Nernst equation.

Answer 33

E = EΘ + RT/zF x ln [oxidised form]/[reduced form] - R = gas constant (8.314 J/K/mol) - z = number of e- transferred in reaction

Question 34

State the simplified Nernst equation.

Answer 34

E = EΘ + 0.059/z x log10 [oxidised form] - at STD temperature, R, T and F are constant. - [reduced] is the metal concentration which doesn't change.

Question 35

What does log10[O] tell us?

Answer 35

It changes the sign of the second term. - if conc. = 1M, log10[O] = 0 and E = EΘ - if conc. <1M, log10[O] is - and E < EΘ - if conc. >1M, log10[O] is + and E > EΘ

Question 36

Describe ordinary dry cells.

Answer 36

They have voltages of 1.5-2.0V. Several are needed to produce enough power, and their voltage drops gradually over time. Cells joined together produce enough voltage but take up space.

Question 37

What must one consider when selecting a cell for a specific function?

Answer 37

- Rechargeability - Size + mass - Maximum voltage - Nature of electrolyte - Cost - How long the cell can deliver its maximum voltage.

Question 38

Describe rechargeable / secondary / storage cells.

Answer 38

Not primary (redox reactions continue until [R] is llow and voltage declines). These are recharged by passing an electric current through, eg. car battery: - uses Pb and PbO2 plates immersed in H2SO4 electrolyte. - 6 x 2V cells = 12V - recharged by the car's alternator when the engine is running.

Question 39

Describe some improved batteries for electric vehicles.

Answer 39

Nickel-cadmium cells - smaller, lower mass, last longer, lower voltage. Aluminium-air cells - lower mass, higher voltage, more expensive, Al anodes occasionally need replacing.

Question 40

Describe solid state cells.

Answer 40

Primary cells developed to produce improved voltage and smaller size and mass. Used in pacemakers, hearing aids, watches, calculators. No liquid/paste = no risk of leakage. Often uses Li or Zn as - pole and iodine / MnO2 / Ag2O as + pole.

Question 41

What are the advantages of other types of cell?

Answer 41

Ni-MH - cheap, high energy density, no toxic materials (Ni-Cd). Li-ion - high energy density, low mass. Both decrease waste, have low long-term cost and save finite/expensive resources.

Question 42

What is a fuel cell?

Answer 42

An electrochemical cell where fuel releases electrons at one electrode and oxygen accepts electrons at the other.

Question 43

How do hydrogen-oxygen fuel cells work?

Answer 43

- H2 and O2 bubbled through two porous platinum-coated electrodes. - H2 is oxidised at the negative electrode, H+ ions diffuse through the proton exchange membrane and electrons flow through the external circuit, releasing energy. - O2 is reduced to water at the positive electrode. - POLE: H2 -> 2H+ + 2e- + POLE: 4H+ + O2 + 4e- -> 2H2O

Question 44

What are the advantages and disadvantages of fuel cells?

Answer 44

ADVANTAGES: - water only product - higher energy/g of fuel compared to petrol engines - efficient - no moving parts where energy could be lost as thermal; direct transfer from cell to motor. DISADVANTAGES: - expensive (electrodes, membrane) - toxic manufacturing byproducts - high pressure tanks to store enough fuel, refuelling needed - H2 is only produced cheaply by using fossil fuels - doesn't work at cold temperatures - cell 'freezes' below zero.

Question 45

Which additional ions are present in an aqueous solution?

Answer 45

H+ and OH- from the ionisation of water.

Question 46

Which ions are discharged during the electrolysis of aqueous solutions?

Answer 46

It depends on the relative E and concentration of ions. - The most easily reduced cation is discharged (eg. NaCl would -> H2 gas). At the graphite ANODE: - the discharge series is followed (increasing ease of oxidation/discharge): - sulfate, nitrate, chloride, hydroxide, bromide, iodide. If conditions are standard, the concentration of the aqueous solution is 1M with respect to the aq. ionic compound. The concentration of hydrogen and hydroxide ions in solution is very low, but as long as the difference in electrode potentials is greater than 0.30V, we can be fairly sure that the predictions will be correct.

Question 47

What happens in the electrolysis of conc. Na2SO4 (aq)?

Answer 47

ANODE: 4OH- -> O2 + 2H2O + 4e- | - OH- discharged, oxidised -> O2.

Question 48

How are electrolysis products affected by solution concentration?

Answer 48

An ion higher in the discharge series may be discharged with preference over one lower if its concentration is much higher than usual. For this to happen, the E value difference would need to be <0.30V. For example, dil. NaCl produces both Cl2 and O2 at the anode; proportion of O2 will increase as dilution increases.

Question 49

How do you draw a cell diagram in IUPAC notation?

Answer 49

Anode | Reductant, Product || Oxidant, Product | Cathode - if either reductant/product is the anode, it is omitted - if either oxidant/product is the cathode, it is omitted.

Question 50

Why are salt bridges used in electrolysis cells?

Answer 50

To complete the circuit and maintain ionic balance between the half-cells. Ions can use this to travel between solutions so there isn't a charge buildup in either half-cell.